ScienceCBSE Class 10 · Electricity

Resistance and Resistivity: Factors Affecting Resistance

Updated 14 October 2026 · Fact-checked

Resistance is how strongly a conductor opposes current. It increases with length, decreases with area of cross-section, and depends on the material and temperature. Resistivity is the material's own property, linked by R = ρl/A, with SI unit ohm metre. The board asks definitions, reasons for material choices and ratio numericals on cut or stretched wires.

At the same voltage, doubling the length doubles R and halves the current; doubling the area halves R and doubles the current, as R = ρl/A.

Illustration drawn for QuestionBuddy.

Understand Resistance and Resistivity

Electrons in a wire are not completely free. The atoms around them pull on them and slow them down. This opposition to the flow of charge is the resistance of the conductor. From Ohm's law, $R = \frac{V}{I}$ , so a larger resistance means a smaller current for the same voltage.

The SI unit of resistance is the ohm ( $Ω$ ). A conductor has a resistance of 1 Ω when a potential difference of 1 V across its ends makes a current of 1 A flow through it, so $1 Ω = \frac{1 V}{1 A}$ . Ohm's law, and so a fixed value of $\frac{V}{I}$ for a wire, holds only while its temperature stays the same.

Activity 11.3 shows what resistance depends on. A nichrome wire is connected to a cell and an ammeter. When a nichrome wire of double length is used, the current falls to half, so the resistance has doubled. When a thicker nichrome wire of the same length is used, the current rises, so the resistance has fallen. When a copper wire of the same length and thickness replaces nichrome, the current changes, so the material matters too.

Why? A longer wire gives electrons a longer path full of obstacles, so $R \propto l$ . A thicker wire is like a wider road: more electrons can pass at once, so $R \propto \frac{1}{A}$ . Combining these, $R = ρ \frac{l}{A}$ . The constant $ρ$ (rho) is the resistivity of the material. Its SI unit is $Ω m$ .

The key difference: resistance belongs to a particular piece of wire, while resistivity belongs to the material. Cut a copper wire in half and its resistance halves, but its resistivity stays the same. Resistivity changes only if the material or the temperature changes. Both resistance and resistivity vary with temperature.

Metals and alloys have very low resistivity, roughly $1 0^{- 8}$ to $1 0^{- 6} Ω m$ , which is why they conduct well. Glass and rubber resist current so strongly that their resistivity is about $1 0^{12}$ to $1 0^{17} Ω m$ . An alloy usually has a higher resistivity than the metals it is made from, and it does not catch fire or corrode easily when red hot, so nichrome is chosen for the elements of irons and toasters. Copper and aluminium, with their low resistivity, carry electricity over long transmission lines. Tungsten is tough and melts only at 3380 °C, so a tungsten filament can glow white-hot without melting.

Draw and label: the Activity 11.3 circuit (Fig. 11.5)

Draw one closed loop with a cell, a plug key, an ammeter and a gap for the wire under test, all in series.
Use the circuit symbols: the cell as a long thin line and a short thick line, and the ammeter as a circle with A inside.
Label the gap and list the four wires placed in it one at a time: (1) nichrome, length l; (2) nichrome, same thickness, length 2l; (3) thicker nichrome, length l; (4) copper, same length and area as (1).
Draw an arrow showing the current going from the positive terminal of the cell, through the circuit, back to the negative terminal.
Common slip: drawing the ammeter in parallel with the wire. It must be in series.

This topic links Ohm's law to circuits: once you can find R from the wire's size and material, you can use it in series, parallel, heating and power calculations.

Resistance and Resistivity: formulas and key facts

Resistance of a uniform conductor: $R = ρ \frac{l}{A}$; l in m, A in $m^{2}$ , $ρ$ in $Ω m$ , R in $Ω$ .
Resistivity: $ρ = \frac{R A}{l}$; Depends only on material and temperature, not on the size or shape of the wire.
Area of a round wire: $A = \frac{π d ^{2}}{4} = π r^{2}$; Area goes as diameter squared: double the diameter and area becomes 4 times. Convert mm to m first.
Proportionality facts: $R \propto l and R \propto \frac{1}{A}$; Double the length, R doubles; double the area, R halves.
Stretched wire (same volume): $R_{2} = n^{2} R_{1}$; If a wire is stretched to n times its length, volume stays fixed, so area becomes $\frac{1}{n}$ and R becomes $n^{2}$ times. Show this using $R = ρ \frac{l}{A}$ .
Resistivity ranges: $ρ_{metals, alloys} \sim 1 0^{- 8} to 1 0^{- 6} Ω m$; Insulators such as glass and rubber: about 10¹² to 10¹⁷ Ω m. Alloys have higher resistivity than their constituent metals.

How to solve Resistance and Resistivity questions

Almost every resistance-resistivity question is either a direct substitution in $R = ρ \frac{l}{A}$ or a ratio between two wires. Use this method:

1Write down the given values with units and decide what is asked: R, $ρ$ , l or A.
2Convert to SI: mm to m (1 mm = $1 0^{- 3}$ m), $m m^{2}$ to $m^{2}$ (1 $m m^{2} = 1 0^{- 6} m^{2}$ ).
3If diameter or radius is given, find area using $A = \frac{π d ^{2}}{4}$ .
4For a single wire, write $R = ρ \frac{l}{A}$ , rearrange, substitute with units and give the answer with unit.
5For two wires of the same material, write the formula for each and divide; $ρ$ cancels.
6If a wire is stretched or melted and redrawn, use constant volume: $l_{1} A_{1} = l_{2} A_{2}$ .
7Check: resistivity never changes for the same material at the same temperature; a longer or thinner wire must give a larger R.

Resistance and Resistivity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Factors affecting resistance and resistivity

Imagine your school lab assistant has a nichrome wire of resistance 8 Ω for a practical. A student needs a 2 Ω wire but there is no other nichrome wire. He cuts the wire into two equal halves, lays them side by side and connects their ends together at both ends, so the current splits between them.

Find the resistance of this arrangement using $R = ρ \frac{l}{A}$ , and check your answer using the rule for resistors in parallel. Does the resistivity change?

Try it first, then show the solutionSolution

Each half has length $\frac{l}{2}$ and area A, so each half has resistance $\frac{8 Ω}{2} = 4 Ω$ .
Joined at both ends, the two halves act like one wire of length $l_{2} = \frac{l}{2}$ and area $A_{2} = 2 A$ .
$R_{2} = ρ \frac{l /2}{2 A} = \frac{1}{4} ρ \frac{l}{A} = \frac{R _{1}}{4}$ .
$R_{2} = \frac{8 Ω}{4} = 2 Ω$ .
Check: two 4 Ω halves in parallel give $\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ , so $R_{p} = 2 Ω$ .
Material and temperature are unchanged, so resistivity stays the same.

Answer: The arrangement has resistance 2 Ω, exactly what the student needed; resistivity is unchanged.

Memory hook: Cut in two, join side by side: half the length, double the area, one quarter the resistance.

Factors affecting resistance and resistivity

Suppose an electrician at your home is choosing wire for the house wiring and for a new room heater element. He has copper wire and nichrome wire of the same size.

Which wire should go in the walls and which in the heater element? Give two reasons for the heater choice. Using resistivities of $1.62 \times 1 0^{- 8} Ω m$ for copper and $100 \times 1 0^{- 6} Ω m$ for nichrome, about how many times larger is the resistance of the nichrome wire than that of a copper wire of the same length and thickness?

Try it first, then show the solutionSolution

Copper has very low resistivity, so little energy is wasted as heat in the wiring.
So copper is used for the house wiring.
Nichrome is an alloy with much higher resistivity, so it produces a lot of heat for the same current.
Nichrome does not oxidise (burn) readily at high temperature, so the element lasts.
Same length and area, so $\frac{R _{ni c h r o m e}}{R _{co pp er}} = \frac{ρ _{ni c h r o m e}}{ρ _{co pp er}} = \frac{100 \times 1 0 ^{- 6}}{1.62 \times 1 0 ^{- 8}} \approx 6200$ .

Answer: Copper for house wiring; nichrome for the heater element because of its high resistivity and because it does not oxidise readily when hot. The nichrome wire has about 6200 times the resistance of a copper wire of the same size.

Memory hook: Copper carries, nichrome heats.

Tricks to remember Resistance and Resistivity

Short memory aids. Say each one aloud once, then use it on a question the same day.

The road picture
Longer road = more traffic jams (more R). Wider road = smoother flow (less R). Different road surface = different material (different ρ).
Use it when: Recall quickly how length, area and material affect R.
Resistance belongs to the wire, resistivity to the material
Think 'R for Rod, ρ for Recipe'. Change the rod's size and R changes; only a new recipe (material) or temperature changes ρ.
Use it when: Spot traps in MCQs about cut, folded or stretched wires.
Stretch rule n squared
Stretch n times, resistance n squared times: stretch 2, R × 4; stretch 3, R × 9.
Use it when: Fast check for stretched-wire questions; still show working in a long answer.

Resistance and Resistivity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Doubling length and tripling area · 2 marks · Very short answerBoard level

A wire of length l and area of cross-section A has resistance 12 Ω. Another wire of the same material has length 2l and area of cross-section 3A. Find its resistance.

Try it first, then show the solutionSolution

$R_{1} = ρ \frac{l}{A} = 12 Ω$ (½ mark)
$R_{2} = ρ \frac{2 l}{3 A} = \frac{2}{3} ρ \frac{l}{A}$ (½ mark)
$R_{2} = \frac{2}{3} \times 12 Ω$ (½ mark)
$R_{2} = 8 Ω$ (½ mark)

Answer: 8 Ω

Example 2: Wire cut into unequal parts: series vs parallel · 3 marks · Short answerHarder board level

A uniform wire of resistance 18 Ω is cut into two pieces whose lengths are in the ratio 1 : 2. (a) Find the resistance of each piece. (b) Find the equivalent resistance when the two pieces are joined in series and when they are joined in parallel. (c) Find the ratio of the series value to the parallel value.

Try it first, then show the solutionSolution

(a) Same material and same area, so $R \propto l$ : the pieces are $\frac{1}{3} \times 18 = 6 Ω$ and $\frac{2}{3} \times 18 = 12 Ω$ (1 mark)
(b) Series: $R_{s} = 6 Ω + 12 Ω = 18 Ω$ (½ mark)
Parallel: $\frac{1}{R _{p}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12}$ , so $R_{p} = 4 Ω$ (1 mark)
(c) $R_{s} : R_{p} = 18 : 4 = 9 : 2$ (½ mark)

Answer: Pieces 6 Ω and 12 Ω; series 18 Ω, parallel 4 Ω; ratio 9 : 2. Series gives back the original resistance; parallel gives less than the smaller piece.

Example 3: Stretched wire · 3 marks · Short answerHarder board level

A wire of resistance 5 Ω is drawn out uniformly so that its length becomes three times the original. Show that the resistance becomes nine times and find the new resistance. What happens to its resistivity?

Try it first, then show the solutionSolution

Volume stays the same: $l_{1} A_{1} = l_{2} A_{2}$ ; with $l_{2} = 3 l_{1}$ , $A_{2} = \frac{A _{1}}{3}$ (1 mark)
$R_{2} = ρ \frac{3 l _{1}}{A _{1} /3} = 9 ρ \frac{l _{1}}{A _{1}} = 9 R_{1}$ (1 mark)
$R_{2} = 9 \times 5 = 45 Ω$ (½ mark)
Resistivity is unchanged as material and temperature are the same (½ mark)

Answer: New resistance = 45 Ω; resistivity unchanged.

Example 4: Diameter in mm: find the material · 3 marks · Short answerHarder board level

A wire 3.14 m long and 0.2 mm in diameter has a resistance of 1.62 Ω. (a) Find its resistivity and use the values copper $1.62 \times 1 0^{- 8} Ω m$ , aluminium $2.63 \times 1 0^{- 8} Ω m$ and nichrome $100 \times 1 0^{- 6} Ω m$ to name the material. (b) Find the resistance of a wire of the same material and length but 0.4 mm in diameter. (Take $π = 3.14$ .)

Try it first, then show the solutionSolution

Radius $r = 0.1$ mm $= 1 \times 1 0^{- 4}$ m, so $A = π r^{2} = 3.14 \times 1 0^{- 8} m^{2}$ . (1 mark)
$ρ = \frac{R A}{l} = \frac{1.62 \times 3.14 \times 1 0 ^{- 8}}{3.14} = 1.62 \times 1 0^{- 8} Ω m$ , so the wire is copper. (1 mark)
(b) Doubling the diameter makes the area 4 times, so $R = \frac{1.62}{4} = 0.405 Ω$ . (1 mark)

Answer: (a) $1.62 \times 1 0^{- 8} Ω m$ , copper. (b) 0.405 Ω (one quarter, because the area is four times).

Example 5: Activity 11.3 explained with a full answer · 5 marks · Long answerHarder board level

(a) Describe an activity to show that the resistance of a conductor depends on its length, area of cross-section and material. (b) Write the relation between R, l and A, and define resistivity with its SI unit. (c) A wire 2 m long with area of cross-section $1 \times 1 0^{- 6} m^{2}$ has resistance 200 Ω. Find the resistivity and name the material using: nichrome $100 \times 1 0^{- 6} Ω m$ , manganin $44 \times 1 0^{- 6} Ω m$ , copper $1.62 \times 1 0^{- 8} Ω m$ .

Try it first, then show the solutionSolution

(a) Connect a cell, an ammeter, a plug key and a nichrome wire of length l in series; note the current (½ mark)
Replace with a nichrome wire of the same thickness but length 2l: current halves, so R doubles (½ mark)
Use a thicker nichrome wire of length l: current increases, so R decreases with area (½ mark)
Use a copper wire of the same length and area: current changes, so R depends on material (½ mark)
(b) $R \propto l$ and $R \propto \frac{1}{A}$ , so $R = ρ \frac{l}{A}$ (½ mark)
Resistivity is the constant $ρ$ , a property of the material; SI unit $Ω m$ (½ mark)
(c) $ρ = \frac{R A}{l} = \frac{200 \times 1 \times 1 0 ^{- 6}}{2}$ (1 mark)
$ρ = 100 \times 1 0^{- 6} Ω m$ , so the material is nichrome (1 mark)

Answer: R depends on length, area and material; $R = ρ \frac{l}{A}$ , unit of ρ is Ω m; ρ = $1 \times 1 0^{- 4} Ω m$ , which is nichrome.

Example 6: Stretched wire in parallel with an unstretched wire · 5 marks · Long answerHarder board level

A nichrome wire of resistance 4 Ω is stretched uniformly until its length is doubled. It is then joined in parallel with an identical unstretched nichrome wire of 4 Ω, and the combination is connected across an 8 V battery. Ignore the resistance of the connecting wires. Find (a) the resistance of the stretched wire, (b) the equivalent resistance of the combination, (c) the total current drawn from the battery, and (d) the potential difference across each wire and the current through each.

Try it first, then show the solutionSolution

(a) Volume stays the same: $l_{1} A_{1} = l_{2} A_{2}$ ; with $l_{2} = 2 l_{1}$ , $A_{2} = \frac{A _{1}}{2}$ (½ mark)
$R_{2} = ρ \frac{2 l _{1}}{A _{1} /2} = 4 ρ \frac{l _{1}}{A _{1}} = 4 \times 4 Ω = 16 Ω$ (1 mark)
(b) $\frac{1}{R _{p}} = \frac{1}{16} + \frac{1}{4} = \frac{1 + 4}{16} = \frac{5}{16}$ (½ mark)
$R_{p} = \frac{16}{5} = 3.2 Ω$ (½ mark)
(c) $I = \frac{V}{R _{p}} = \frac{8 V}{3.2 Ω} = 2.5 A$ (1 mark)
(d) In parallel, the potential difference across each wire equals that across the combination: 8 V (½ mark)
Stretched wire: $I_{1} = \frac{8 V}{16 Ω} = 0.5 A$ ; unstretched wire: $I_{2} = \frac{8 V}{4 Ω} = 2 A$ (½ mark)
Check: $I_{1} + I_{2} = 0.5 A + 2 A = 2.5 A$ , equal to the total current (½ mark)

Answer: (a) 16 Ω (b) 3.2 Ω (c) 2.5 A (d) 8 V across each wire; 0.5 A in the stretched wire and 2 A in the unstretched wire.

Example 7: Case-based: choosing wires · 4 marks · Case-basedBoard level

Read the passage and answer.

A student measures two wires P and Q, both of length 1 m and area of cross-section $1 \times 1 0^{- 6} m^{2}$ , at the same temperature. P has resistance 0.0162 Ω and Q has resistance 100 Ω. She then cuts wire Q into two equal halves.

(a) Find the resistivity of P and of Q. (1 mark)
(b) Which wire would you use for a toaster element and which for connecting wires? Give a reason for each. (2 marks)
(c) What is the resistance and the resistivity of each half of Q? (1 mark)

Try it first, then show the solutionSolution

(a) $ρ = \frac{R A}{l}$ : P gives $1.62 \times 1 0^{- 8} Ω m$ , Q gives $1 \times 1 0^{- 4} Ω m$ (1 mark)
(b) Q for the toaster: very high resistivity, so it produces much heat; such an alloy does not oxidise readily when hot (1 mark)
P for connecting wires: very low resistivity, so little energy is lost as heat (1 mark)
(c) Halving length halves R: each half is 50 Ω; resistivity stays $1 \times 1 0^{- 4} Ω m$ (1 mark)

Answer: (a) P: $1.62 \times 1 0^{- 8} Ω m$ ; Q: $1 \times 1 0^{- 4} Ω m$ . (b) Q for the toaster, P for connecting wires. (c) 50 Ω each, resistivity unchanged.

Where students lose marks in Resistance and Resistivity

The slips examiners see most often, and what to write instead.

Mistake: Saying resistivity changes when the wire is cut or stretched
Students mix up resistance (a property of the wire) with resistivity (a property of the material).
Wrong working
Wire stretched to double length, so resistivity doubles.
Right working
Resistance becomes 4 times; resistivity is unchanged because the material and temperature are the same.
Do this: Ask: has the material or temperature changed? If not, ρ is the same.
Usually costs: The full 1 mark in MCQs and A-R; ½ to 1 mark in reasoning answers.
Mistake: Doubling area when diameter doubles
Forgetting that area depends on the square of the diameter.
Wrong working
Diameter doubled, so A doubles and R becomes half.
Right working
$A = \frac{π d ^{2}}{4}$ : diameter doubled gives A four times, so R becomes one quarter.
Do this: Always write $A \propto d^{2}$ before comparing wires.
Usually costs: The final answer mark and often the step mark.
Mistake: Not converting mm to m
Students substitute 0.5 mm as 0.5 in the formula.
Wrong working
$A = \frac{3.14 \times 0. 5 ^{2}}{4}$ used with ρ in Ω m.
Right working
$d = 0.5 \times 1 0^{- 3} m$ , so $A = \frac{3.14 \times 0.25 \times 1 0 ^{- 6}}{4} m^{2}$ .
Do this: Convert every length to metres in the 'given' line before any working.
Usually costs: 1 to 1½ marks in a 3-mark numerical.
Mistake: Ignoring the change in area for a stretched wire
Students change only the length and forget volume is constant.
Wrong working
Length tripled, so R tripled.
Right working
Length tripled, area becomes one third, so R becomes 9 times.
Do this: Write $l_{1} A_{1} = l_{2} A_{2}$ first in every stretching question.
Usually costs: Most of the marks in the question.

Exam tips

In ratio questions, write $R = ρ \frac{l}{A}$ for both wires and divide; examiners award a mark for showing ρ cancelling.
For 'give reasons' on alloys, mention both points: high resistivity and no oxidation at high temperature.
Assertion-reason items often test whether a size change alters resistivity; judge A and R separately first.
When describing Activity 11.3, say what is changed, what is kept the same and what the ammeter shows each time.
Always give the unit Ω m for resistivity; writing Ω or Ω/m loses the mark.

Practice questions on Resistance and Resistivity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A wire of resistance R is stretched uniformly so that its length becomes twice its original length. Its new resistance is:
(a) $\frac{R}{2}$
(b) 2R
(c) 4R
(d) $\frac{R}{4}$
Hint
The volume stays the same, so the area changes too.
Answer
(c) 4R. Length doubles and area halves, so $R_{2} = ρ \frac{2 l}{A /2} = 4 R$ .
Q2 · 3 marks · Short answerHarder board level
Two wires A and B are of the same material. A has length l and diameter d; B has length 2l and diameter 2d. Find the ratio of their resistances $R_{A} : R_{B}$ .
Hint
Area goes as diameter squared; divide the two formulas.
Answer
$A_{B} = 4 A_{A}$ . $\frac{R _{A}}{R _{B}} = \frac{l / A _{A}}{2 l /4 A _{A}} = 2$ . So $R_{A} : R_{B} = 2 : 1$ .
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): A thick wire of a metal carries more current than a thin wire of the same metal and length when connected to the same source.
Reason (R): The resistivity of a thick wire is less than that of a thin wire of the same metal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Does thickness change resistivity or resistance?
Answer
(c) A is true (larger area, lower resistance, more current) but R is false: resistivity depends only on material and temperature.
Q4 · 3 marks · Short answerHarder board level
A wire of resistance 16 Ω is cut into four equal parts and the parts are joined in parallel. (a) Find the resistance of each part. (b) Find the equivalent resistance. (c) Explain, using $R = ρ \frac{l}{A}$ , why the result is the same as one wire of one quarter length and four times area.
Hint
Cut first, then use the parallel rule for equal resistors.
Answer
(a) 4 Ω each. (b) $\frac{1}{R _{p}} = 4 \times \frac{1}{4}$ , so $R_{p} = 1 Ω$ . (c) Length $\frac{l}{4}$ , area 4A gives $ρ \frac{l}{16 A} = \frac{16}{16} = 1 Ω$ , same answer.
Q5 · 3 marks · Short answerHOTS / competency
A student finds that a wire's resistance doubles when she replaces it with another wire. She claims the new wire must be made of a different material. Give two other changes that could explain this, and state what measurement would confirm her claim.
Hint
List every factor in $R = ρ \frac{l}{A}$ and the temperature effect.
Answer
The new wire could be twice as long, or have half the area of cross-section (or be at a different temperature). To confirm a different material, measure R, l and A, calculate $ρ = \frac{R A}{l}$ at the same temperature and compare; a different ρ means a different material.
Q6 · 4 marks · Case-basedHOTS / competency
Read the data and answer.
A student measures four wires at the same temperature:
- W1: l = 1 m, A = $1 \times 1 0^{- 6} m^{2}$ , R = 100 Ω
- W2: l = 2 m, A = $2 \times 1 0^{- 6} m^{2}$ , R = 100 Ω
- W3: l = 1 m, A = $1 \times 1 0^{- 6} m^{2}$ , R = 44 Ω
- W4: l = 0.5 m, A = $0.25 \times 1 0^{- 6} m^{2}$ , R = 88 Ω
Resistivity values: nichrome $100 \times 1 0^{- 6} Ω m$ , manganin $44 \times 1 0^{- 6} Ω m$ .
(a) Calculate the resistivity of each wire. (2 marks)
(b) Which two pairs are made of the same material? Name each material. (1 mark)
(c) W1 and W2 have the same resistance. A classmate says this alone proves they are the same material. Is this right? Justify. (1 mark)
Hint
Find $ρ = \frac{R A}{l}$ for each wire; same material means same ρ at the same temperature.
Answer
(a) $ρ = \frac{R A}{l}$ . W1: $\frac{100 \times 1 \times 1 0 ^{- 6}}{1} = 100 \times 1 0^{- 6} Ω m$ ; W2: $\frac{100 \times 2 \times 1 0 ^{- 6}}{2} = 100 \times 1 0^{- 6} Ω m$ ; W3: $\frac{44 \times 1 \times 1 0 ^{- 6}}{1} = 44 \times 1 0^{- 6} Ω m$ ; W4: $\frac{88 \times 0.25 \times 1 0 ^{- 6}}{0.5} = 44 \times 1 0^{- 6} Ω m$ . (b) W1 and W2 are nichrome; W3 and W4 are manganin. (c) No. Resistance depends on length and area as well as material, so wires of different materials can have equal R if their sizes differ. W1 and W2 are the same material because their resistivities are equal at the same temperature.
Q7 · 3 marks · Short answerHOTS / competency
A student plots V (on the y-axis) against I (on the x-axis) for two wires X and Y made of the same material and of the same length, at the same temperature. Both graphs are straight lines through the origin. The line for X is steeper than the line for Y. At a current of 2 A, the graph shows 8 V for X and 2 V for Y.
(a) Which wire has the greater resistance? Find both resistances. (1 mark)
(b) Which wire is thicker? Find the ratio of their areas of cross-section $A_{X} : A_{Y}$ . (1 mark)
(c) What does a straight line through the origin tell you, and under what condition does it hold? (1 mark)
Hint
On a V-I graph, V/I at any point gives R; a steeper line means a larger R.
Answer
(a) $R = \frac{V}{I}$ : $R_{X} = \frac{8 V}{2 A} = 4 Ω$ , $R_{Y} = \frac{2 V}{2 A} = 1 Ω$ ; X (the steeper line) has the greater resistance. (b) Same material and length, so $R \propto \frac{1}{A}$ : $A_{X} : A_{Y} = R_{Y} : R_{X} = 1 : 4$ , so Y is thicker. (c) V is directly proportional to I, so V/I is constant (Ohm's law); this holds only while the temperature of the wire stays the same.
Q8 · 3 marks · Short answerHarder board level
A wire 1 m long and 0.5 mm in diameter has a resistance of 224 Ω. (a) Find its resistivity (take $π = 3.14$ ) and identify the material from: nichrome $100 \times 1 0^{- 6} Ω m$ , manganin $44 \times 1 0^{- 6} Ω m$ , constantan $49 \times 1 0^{- 6} Ω m$ . (b) What length of this wire is needed to make a 56 Ω resistor?
Hint
Convert the diameter to metres and halve it before using $A = π r^{2}$ .
Answer
(a) $r = 0.25$ mm $= 2.5 \times 1 0^{- 4}$ m, $A = 3.14 \times 6.25 \times 1 0^{- 8} \approx 1.96 \times 1 0^{- 7} m^{2}$ . $ρ = \frac{R A}{l} = 224 \times 1.96 \times 1 0^{- 7} \approx 44 \times 1 0^{- 6} Ω m$ , so the wire is manganin (2 marks). (b) $R \propto l$ for the same wire: $l = \frac{56}{224} \times 1 = 0.25$ m (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Three factors on which resistance depends?Show answer
Length, area of cross-section and nature of material (and it also varies with temperature).
2.Formula linking R, ρ, l and A?Show answer
$R = ρ \frac{l}{A}$
3.SI unit of resistivity?Show answer
Ohm metre ( $Ω m$ )
4.SI unit of resistance and definition of 1 ohm?Show answer
Ohm (Ω). 1 Ω is the resistance of a conductor when a potential difference of 1 V across it makes a current of 1 A flow.
5.Effect on R of doubling the length?Show answer
R doubles.
6.Why are copper and aluminium used in transmission lines?Show answer
They have low resistivity, so little energy is lost as heat.
7.Why is tungsten used for bulb filaments?Show answer
It is a strong metal with a very high melting point (3380 °C), so the filament can get hot enough to give out light without melting.
8.Resistivity range of metals and alloys?Show answer
About $1 0^{- 8}$ to $1 0^{- 6} Ω m$ .

Resistance and Resistivity: frequently asked questions

What is the difference between resistance and resistivity?

Resistance is the opposition offered by a particular conductor and depends on its length, thickness, material and temperature. Resistivity is a property of the material alone and depends only on the material and temperature. Two copper wires of different sizes have different resistances but the same resistivity.

Does the resistivity of a wire change if I cut it?

No. Cutting changes the length, so the resistance changes, but the material is the same, so resistivity stays the same at a given temperature.

Why does a thick wire have less resistance than a thin one?

A thick wire has a larger area of cross-section, so more electrons can flow through it at the same time. Resistance is inversely proportional to area, so a larger area means lower resistance and more current for the same voltage.

Do I need to memorise the resistivity table?

No. The NCERT book says you need not memorise the values; they are given for solving numericals. You should know the ranges for metals and insulators and that alloys have higher resistivity than their metals.

What happens to resistance when a wire is stretched?

Its volume stays the same, so if the length becomes n times, the area becomes 1/n times. Since R = \rho \frac{l}{A}, the resistance becomes n^2 times. The resistivity does not change, because the material and temperature are the same.